Two moles of HCHO and 1 mol of PhCHO react with conc. NaOH. What are the produsts quantitatively?

Explanation

Initially there will be crossed Cannizzaro reaction between HCHO and PhCHO.
`HCHO+PhCHO overset(NaOH)rarr underset(1mol)(HCOONa) + underset(1mol)(PhCH_(2)OH)`
`underset((1/2)mol)(HCHO) + underset((1/2)mol)(HCHO) overset(NaOH)rarr underset((1/2)mol)(HCOONa) + underset((1/2)mol)(CH_(3)OH)`
Remaining 1 mol of HCHO will give Cannizzaro reaction. So `1//2` mol of HCHO will react with `1//2` mol of HCHO to give `1//2` mol of HCOONa and `1//2` mol of `CH_(3)OH`. Hence, `1.5` mol of HCOONa, 1 mol of `PhCH_(2)OH`, and `1//2` mol of `CH_(3)OH` are formed.